One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300K then total entropy change of system in the above process is:
`[R=0.082L atm mol^(-1) K^(-1)-=8.3J mol^(-1) K^(-1)]`

Options

A. 0
B. `R In (24.6)`
C. `R In (2490)`
D. `(3)/(2) R In (24.6)`

Answer

To find the total entropy change of the system during the isothermal expansion of one mole of an ideal monoatomic gas, we can follow these steps:

### Step 1: Understand the Process
The gas expands isothermally (at constant temperature) from an initial volume \( V_i = 1 \, \text{L} \) to a final volume \( V_f \) where its pressure equals the external pressure of \( 1 \, \text{atm} \).

### Step 2: Calculate the Final Volume
Using the ideal gas law, we can find the final volume. The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles (1 mole)
- \( R \) = ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\))
- \( T \) = temperature (in K)

At the final state, the pressure \( P = 1 \, \text{atm} \) and \( T = 300 \, \text{K} \):
\[ V_f = \frac{nRT}{P} = \frac{1 \, \text{mol} \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}}{1 \, \text{atm}} = 24.63 \, \text{L} \]

### Step 3: Calculate the Change in Entropy
The change in entropy \( \Delta S \) for an isothermal process can be calculated using the formula:
\[ \Delta S = nR \ln\left(\frac{V_f}{V_i}\right) \]
Substituting the values:
\[ \Delta S = 1 \, \text{mol} \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times \ln\left(\frac{24.63 \, \text{L}}{1 \, \text{L}}\right) \]
Calculating the logarithm:
\[ \ln\left(24.63\right) \approx 3.20 \]
Now substituting this value back:
\[ \Delta S = 0.0821 \times 3.20 \approx 0.263 \, \text{J K}^{-1} \]

### Step 4: Final Result
The total entropy change of the system during the isothermal expansion is approximately:
\[ \Delta S \approx 0.263 \, \text{J K}^{-1} \]